AMC12 2006 A

AMC12 2006 A · Q23

AMC12 2006 A · Q23. It mainly tests Exponents & radicals, Sequences & recursion (algebra).

Given a finite sequence $S=(a_1,a_2,\ldots ,a_n)$ of $n$ real numbers, let $A(S)$ be the sequence $\left(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}\right)$ of $n-1$ real numbers. Define $A^1(S)=A(S)$ and, for each integer $m$, $2\le m\le n-1$, define $A^m(S)=A(A^{m-1}(S))$. Suppose $x>0$, and let $S=(1,x,x^2,\ldots ,x^{100})$. If $A^{100}(S)=(1/2^{50})$, then what is $x$?

给定由 $n$ 个实数组成的有限序列 $S=(a_1,a_2,\ldots ,a_n)$，令 $A(S)$ 为由 $n-1$ 个实数组成的序列 $\left(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}\right)$。定义 $A^1(S)=A(S)$，并对每个整数 $m$（$2\le m\le n-1$）定义 $A^m(S)=A(A^{m-1}(S))$。设 $x>0$，令 $S=(1,x,x^2,\ldots ,x^{100})$。若 $A^{100}(S)=(1/2^{50})$，求 $x$。

(A) $\frac{1-\sqrt{2}}{2}$ $\frac{1-\sqrt{2}}{2}$

(B) $\sqrt{2}-1$ $\sqrt{2}-1$

(D) $2-\sqrt{2}$ $2-\sqrt{2}$

(E) $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

\[A^1(S)=\left(\frac{1+x}{2},\frac{x+x^2}{2},...,\frac{x^{99}+x^{100}}{2}\right)\] \[A^2(S)=\left(\frac{1+2x+x^2}{2^2},\frac{x+2x^2+x^3}{2^2},...,\frac{x^{98}+2x^{99}+x^{100}}{2^2}\right)\] \[\Rightarrow A^2(S)=\left(\frac{(x+1)^2}{2^2},\frac{x(x+1)^2}{2^2},...,\frac{x^{98}(x+1)^2}{2^2}\right)\] In general, $A^n(S)=\left(\frac{(x+1)^n}{2^n},\frac{x(x+1)^n}{2^n},...,\frac{x^{100-n}(x+1)^n}{2^n}\right)$ such that $A^n(s)$ has $101-n$ terms. Specifically, $A^{100}(S)=\frac{(x+1)^{100}}{2^{100}}$ To find x, we need only solve the equation $\frac{(x+1)^{100}}{2^{100}}=\frac{1}{2^{50}}$. Algebra yields $x=\sqrt{2}-1$.

\[A^1(S)=\left(\frac{1+x}{2},\frac{x+x^2}{2},...,\frac{x^{99}+x^{100}}{2}\right)\] \[A^2(S)=\left(\frac{1+2x+x^2}{2^2},\frac{x+2x^2+x^3}{2^2},...,\frac{x^{98}+2x^{99}+x^{100}}{2^2}\right)\] \[\Rightarrow A^2(S)=\left(\frac{(x+1)^2}{2^2},\frac{x(x+1)^2}{2^2},...,\frac{x^{98}(x+1)^2}{2^2}\right)\] 一般地，$A^n(S)=\left(\frac{(x+1)^n}{2^n},\frac{x(x+1)^n}{2^n},...,\frac{x^{100-n}(x+1)^n}{2^n}\right)$，其中 $A^n(s)$ 有 $101-n$ 项。特别地，$A^{100}(S)=\frac{(x+1)^{100}}{2^{100}}$。要找 $x$，只需解方程 $\frac{(x+1)^{100}}{2^{100}}=\frac{1}{2^{50}}$。代数运算得 $x=\sqrt{2}-1$。

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